The Spetner Anomaly

From EvoWiki

Lee M. Spetner is an information theorist of the “Intelligent Design” variety. In an interchange with Edward E. Max (starting with Max’ TalkOrigins article The Evolution of Improved Fitness) Spetner has tried to state why mutations cannot increase information. The tricky bit here is that the way Spetner defines information increase/decrease leads to some (possibly) weird results. But let’s look at, what we are dealing with, shall we? Spetner’s argumentation can be found in Lee Spetner/Edward Max Dialogue and in How is information content measured?.

Spetner operates with reified information, that is, he treats information as a thing that exists independent of a human observer:

The information content of the genome is difficult to evaluate with any precision. Fortunately, for my purposes, I need only consider the change in the information in an enzyme caused by a mutation. The information content of an enzyme is the sum of many parts, among which are:

• Level of catalytic activity
• Specificity with respect to the substrate
• Strength of binding to cell structure
• Specificity of binding to cell structure
• Specificity of the amino-acid sequence devoted to specifying the enzyme for degradation

These are all difficult to evaluate, but the easiest to get a handle on is the information in the substrate specificity.

We may study bacteria and see how their enzymes catalyze various substrates. Hereby we generate information, but as Spetner has it, we do not generate information, we discover it, so therefore he treats the information as something predefined. And in this case it’s the change in information content of an enzyme and a mutated variant of the same enzyme. This in order to prove that mutations cannot increase information.

As Spetner admits, given this view of information, it’s no easy task to estimate the information content of an enzyme. But, no reason to panic:

To estimate the information in an enzyme I shall assume that the information content of the enzyme itself is at least the maximum information gained in transforming the substrate distribution into the product distribution. (I think this assumption is reasonable, but to be rigorous it should really be proved.) We can think of the substrate specificity of the enzyme as a kind of filter. The entropy of the ensemble of substances separated after filtration is less than the entropy of the original ensemble of the mixture. We can therefore say that the filtration process results in an information gain equal to the decrease in entropy.

Here things start to get too weird for comfort. The main thing here is of course that the “information content” of an enzyme is defined in a way such that what is really measured (to the extent that anything is measured) is the differential efficiency of the enzyme for different sugars.

Spetner continues with a thought experiment:

Let’s imagine a uniform distribution of substrates presented to many copies of an enzyme. I choose a uniform distribution of substrates because that will permit the enzyme to express its maximum information gain. The substrates considered here are restricted to a set of similar molecules on which the enzyme has the same metabolic effect. This restriction not only simplifies our exercise but it applies to the case I discussed in my book.

The products of a substrate on which the enzyme has a higher activity will be more numerous than those of a substrate on which the enzyme has a lower activity. Because of the filtering, the distribution of concentrations of products will have a lower entropy than that of substrates. Note that we are neglecting whatever entropy change stems from the chemical changes of the substrates into products, and we are focusing on the entropy change reflected in the distributions of the products of the substrates acted upon by the enzyme.

Again, what we are really measuring is the efficiency of the enzyme, which can be measured by an experiment as sketched by Spetner. But we are not measuring information, we are generating information by performing the experiment.

Let  {x_i}_{i= 1,…,n}  be the letters occurring in some string S and let f_i be the frequency of x_i. In Shannon information theory, the information associated with x_i, i = 1, …, n is defined by</p>

I(x_i) = -log₂(f_i), i = 1, …, n

This is also called the surprisal value of x_i. The lower the frequency, the higher the surprise.

The entropy of S = the average information of the letters is defined by

(1) H(S) = Σ_i=1,…,n f_i I(x_i) = - Σ_i=1,…,n f_i log₂(f_i)

The binary logarithm is used because the simplest question you can ask with a definite answer is a yes/no-question. The entropy can be interpreted as the maximum minimum number of yes/no-questions you need to ask to find out, what the i ^th letter is. Basically the entropy is a measure of uncertainty; the lower the entropy, the fewer questions you need to ask. The entropy function assumes its maximum (log₂(n)) for f_i = 1/n for all i = 1, …, n.

The formula H(S) = - Σ_i=1,…,n f_i log₂(f_i) for entropy is the same as Spetner’s formula (1), except that Spetner omits the ‘-’ in front of the summation-sign.

In Spetner’s case, the letters  {x_i}_{i =1,…,n</span>}  are the various substrates, and the string S is the ensemble of substrates.

Let’s return to Spetner:

As a first illustration of this formula let us take the extreme case where there are n possible substrates, and the enzyme has a nonzero activity on only one of them. This is perfect filtering. The input entropy for a uniform distribution of n elements is, from (1), given by

(2) H_I = log n

since the f_i’s are each ¹/_n. The entropy of the output is zero,

(3) H_O = 0

because all the concentrations except one are zero, and the concentration of that one is 1. Then the decrease in entropy brought about by the selectivity of the enzyme is then the difference between (2) and (3), or

H = H_I - H_O = log n

Ok, there are THREE obvious flaws here. First, n is a parameter we have selected. How can that have been encoded in the enzyme? In particular, let n = 1. Then H_I = H_O. Second, what if the enzyme cannot catalyse any of the substrates? Then H_O = 0 log 0 = 0 (by definition), which is the same result as with one output product. Third, why don’t we instead calculate the entropy of the substrates that have not been catalysed? Then we would be able to distinguish between those two cases.

Back to Spetner:

Another example is the other extreme case in which the enzyme does not discriminate at all among the n substrates. In this case the input and output entropies are the same, namely

(4) H_I = H_O = log n

Therefore, the information gain, which is the difference between H_O and H_I, in this case is zero,

(5) H = 0.

From this Spetner goes on with discussing a case, where experiments had been done with “wild” bacteria and a mutant version of the same bacteria and various sugars (= Spetner’s substrates). In these experiments the “wild” bacteria exhibited larger specificity for one sugar than the mutant bacteria. From the above, Spetner concludes that mutations cannot increase information.

Well, if you want a certain result, there’s always a way to define you to it, isn’t there? However, while we are in experimentation mood, let’s do some more experimentation. Let’s assume we have two students A and B that are to take an exam. There are n examination questions. Student A prepares for only one of the questions, such that she will get a very good grad, if she is given that question, but a very low grade, if she is given any other question – that is, student A corresponds with the “wild” bacteria. Student B prepares for all n questions, such that he will get a descent grade, no matter which question he is given, not a very good grade for any question, but neither a very low grade for any question – that is, student B corresponds with the mutant bacteria. Which student is most likely to pass the exam? Well, student B will pass the exam independent of the question given, whereas student A will only pass the exam, if she is given that one question she has prepared for. If n = 1, no problem; but let n = 10, 20, 100, …. Which of these two students would represent most information? The one that can only answer one question, or the one that can answer all of them?

Occasionally, specialization is advantageous, occasionally it isn’t. In Spetner’s world (if he were consistent), you would count as most informed, if you knew only one thing, but you knew it better than anybody else. That would be ok, if nobody asked you about anything else, and you never spoke about anything else. But what are the chances for this happening?

Related Articles

• The Dembski Anomaly

External Links

• Entropy, Disorder and Life, excellent TalkOrigins article by John Pieper explaining, why the 2nd law of thermodynamics is not relevant for biological evolution.